Difference between revisions of "2010 AIME II Problems/Problem 12"
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− | <math>\begin{array}{ | + | <math>\begin{array}{cccl} |
7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ | 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ | ||
− | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ | + | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{(b+8c)(b-8c)})&{}\\ |
− | 7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that} a+7c=b+8c)\\ | + | 7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\\ |
49a-343c&=&64b-512c&{}\\ | 49a-343c&=&64b-512c&{}\\ | ||
49a+169c&=&64b&{}\\ | 49a+169c&=&64b&{}\\ | ||
− | 49a+169(a-b)&=&64b&\text{(Note that} c=a-b)\\ | + | 49a+169(a-b)&=&64b&\text{(Note that } c=a-b)\\ |
218a&=&233b&{}\\ | 218a&=&233b&{}\\ | ||
\end{array}</math> | \end{array}</math> | ||
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</center> | </center> | ||
Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>. | Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>. | ||
+ | |||
+ | == Solution 3== | ||
+ | |||
+ | Let the first triangle have sides <math>16n,a,a</math>, so the second has sides <math>14n,a+n,a+n</math>. The height of the first triangle is <math>\frac{7}{8}</math> the height of the second triangle. Therefore, we have <cmath>a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).</cmath> Multiplying this, we get <cmath>64a^2-4096n^2=49a^2+98an-2352n^2,</cmath> which simplifies to <cmath>15a^2-98an-1744n^2=0.</cmath> Solving this for <math>a</math>, we get <math>a=n\cdot\frac{218}{15}</math>, so <math>n=15</math> and <math>a=218</math> and the perimeter is <math>15\cdot16+218+218=\boxed{676}</math>. | ||
+ | |||
+ | ~john0512 | ||
== See also == | == See also == | ||
+ | Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o | ||
{{AIME box|year=2010|num-b=11|num-a=13|n=II}} | {{AIME box|year=2010|num-b=11|num-a=13|n=II}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:24, 25 November 2021
Problem
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is . Find the minimum possible value of their common perimeter.
Solution 1
Let the first triangle have side lengths , , , and the second triangle have side lengths , , , where .
Equal perimeter:
Equal Area:
Since and are integer, the minimum occurs when , , and . Hence, the perimeter is .
Solution 2
Let be the semiperimeter of the two triangles. Also, let the base of the longer triangle be and the base of the shorter triangle be for some arbitrary factor . Then, the dimensions of the two triangles must be and . By Heron's Formula, we have
Since and are coprime, to minimize, we must have and . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by , which gives us a final answer of .
Solution 3
Let the first triangle have sides , so the second has sides . The height of the first triangle is the height of the second triangle. Therefore, we have Multiplying this, we get which simplifies to Solving this for , we get , so and and the perimeter is .
~john0512
See also
Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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